STABILITY OF NEGATIVE STIFFNESS VISCOELASTIC SYSTEMSsilver.neep.wisc.edu/~lakes/NegStfQAM05.pdf · with extreme high stiﬀness analyzed here was metastable. We established an explicit

QUARTERLY OF APPLIED MATHEMATICS

VOLUME LXIII, NUMBER 1

MARCH 2005, PAGES 34–55

S 0033-569X(04)00938-6

Article electronically published on December 17, 2004

STABILITY OF NEGATIVE STIFFNESS VISCOELASTIC SYSTEMS

By

YUN-CHE WANG (Department of Engineering Physics, Engineering Mechanics Program, Universityof Wisconsin-Madison, 147 Engineering Research Building, 1500 Engineering Drive, Madison, WI

53706-1687)

and

RODERIC LAKES (Department of Engineering Physics, Engineering Mechanics Program,Biomedical Engineering Department; Materials Science Program and Rheology Research Center,University of Wisconsin-Madison, 147 Engineering Research Building, 1500 Engineering Drive,

Madison, WI 53706-1687)

Abstract. We analytically investigate the stability of a discrete viscoelastic systemwith negative stiffness elements both in the time and frequency domains. Parametricanalysis was performed by tuning both the amount of negative stiffness in a standardlinear solid and driving frequency. Stability conditions were derived from the analyti-cal solutions of the differential governing equations and the Lyapunov stability theorem.High frequency response of the system is studied. Stability of singularities in the dissipa-tion tan δ is discussed. It was found that stable singular tan δ is achievable. The systemwith extreme high stiffness analyzed here was metastable. We established an explicitlink for the divergent rates of the metastable system between the solutions of differentialgoverning equations in the time domain and the Lyapunov theorem.

1. Nomenclature.

M , m1, m2: mass.K, k1, k2: stiffness for positive stiffness elements.c, c1, c2: damping coefficient.α, α1, α2: ratio between the two spring elements in a standard linear solid. α = α1 whenα2 = 0.γ1, γ2: ratio of damping coefficient of a standard linear solid to that of a damper parallel-connected to the standard linear solid, i.e., γ = η/c.δ: phase angle. Define tan δ = (k∗)/(k∗), where k* denotes the dynamic complexspring constant.κ1, κ2: ratio of stiffness of the series spring in a standard linear solid to that of a springparallel-connected to the standard linear solid, i.e., κ = y/k.

Received December 15, 2003.2000 Mathematics Subject Classification. Primary 74B10; Secondary 74C10, 74D05.E-mail address: [email protected]

c©2004 Brown University

34

STABILITY OF NEGATIVE STIFFNESS VISCOELASTIC SYSTEMS 35

km

u, F

Fig. 1. Spring-mass system, one degree of freedom (dof).

y, y1, y2, z, z1, z2: stiffness for standard linear solid elements, to be tuned. y1 = κ1k1,y2 = κ2k2, y = y1, z = z1.η, η1, η2: damping coefficient for standard linear solids. η1 = γ1c1, η2 = γ2c2, η = η1.u, u1, u2: displacement coordinate of a node.u : du

dt .F , F1, F2: applied force at a node.ω, Ω: frequency.H : Hamiltonian.D: differential operator, d

dt .h: superscript, homogeneous solutions of a differential equation.p: superscript, particular solutions of a differential equation.subscripts: labels of components in a model.

2. Introduction. A bulk solid object of negative stiffness materials is not stable.However, it might be possible to create a stable configuration for composites with a neg-ative stiffness component embedded. Extreme material properties and stability-relatedissues have been reported in [3]–[6] and [11]–[14]. Here, we first review the stability of asimple discrete mechanical model, a spring-mass system, to mathematically demonstrateits internal stability [9] and the stability under external excitations. Then we investigatethe stability of systems with negative stiffness components. This single degree of freedom(dof) system is shown in Figure 1.

By Newton’s second law, the equation of motion is

mu + ku = F, (1)

where m is the mass, k the stiffness, u = u(t) the displacement, and F = F (t) the externaldriving force. A superdot indicates the derivative with respect to time. The solution ofthe equation of motion in the time domain is as follows. Assume F = P + A cosΩt andP and A are constants.

u = uh + up, (2)

where the homogeneous solution uh = C1eλ1t +C2e

λ2t, where λ1,2 = ±iω and ω2 = k/m,and C1 and C2 are constants to be determined by initial conditions. As for the particular

36 YUN-CHE WANG and RODERIC LAKES

solutions, up, one can express them as follows.

up =P

k+

A cosΩt

−mΩ2 + k, when Ω = ω. (3)

up =P

k+

t

2ΩA sin Ωt, when Ω = ω. (4)

It is clear to see that instability occurs only when Ω = ω, due to time-growing be-havior in its particular solution. Therefore, from this time domain analysis, the stabilitycriterion is that the system is stable when Ω = ω. This stability criterion can also beobtained through analysis in the frequency domain. Applying a Fourier transform onEq. (1), one converts the governing equation into an algebraic equation, as follows.

u

F=

1−mω2 + k

. (5)

The instability of the system occurs only when ω2 = k/m, which is consistent withprevious results from Eq. (4). Furthermore, the results indicate that the system isinternally stable when F = 0 and k/m > 0. Moreover, the system is also stable whenk < 0 and m < 0. For k/m < 0, there are no oscillatory solutions for the system, onlyreal exponential ones. Hence, it is internally unstable.

The internal stability is the stability of a system under no external forcing. However,in some cases, it is important to investigate a dynamical system with time variablesexplicitly (i.e., non-autonomous systems), such as flutter analysis. A general method forattacking this problem is to consider the time variable as an ordinary spatial variable [10].However, by doing so, the Lyapunov indirect method will not be suitable for stabilityanalysis since it is local. The Lyapunov direct method can be applied, but stability mustbe checked at all times in the time domain [8].

Alternatively, one can rigorously derive the stability criteria of a mechanical systemby using the so-called extended energy method [7]. Using the system depicted in Eq. (1)as an example, we first rewrite the equation of motion for the system as follows.

q + ω2q =A

mcosΩt (6)

where q = q(t) is the generalized coordinate, ω2 the ratio of k to m, and Ω the drivingfrequency. q(t) is different from u(t) in that the latter contains the contribution from thedead load P . The total energy of the system can be calculated as follows.

E =∫ t

0

(q + ω2q − A

mcosΩt)qdt =

12q2 +

12ω2q2 −

∫ t

0

(A

mcosΩt)qdt. (7)

It is clear to see that the energy is not a first integral of the system. From Eq. (7), itis understood that the Hamiltonian and non-conservative force can be written as follows.

H =12p2 +

12ω2q2, and (8)

N =A

mcosΩt. (9)


Here p = q. Define φ = φ(q, λ, Ω) = dE, and, for E = E(q, p, Ω), the explicitrepresentations of φ and dφ are as follows.

φ =∂H

∂q− N + G = 0, G =

∂H

∂p

dp

dq(10)

dφ =∂φ

∂qdq +

∂φ

∂λdλ +

∂φ

∂ΩdΩ = 0 (11)

or

(∂2H

∂q2− ∂N

∂q+

∂G

∂q)dq + (

∂G

∂λ− ∂N

∂λ)dλ + (

∂2H

∂Ω∂q− ∂N

∂Ω)dΩ = 0. (12)

Here λ is the characteristic frequency of the system. In the present case, λ is chosento be Ω, and Ω is the control parameter in the stability analysis, which means the resultsof the stability analysis will indicate a specific value or region for Ω to make the systemunstable. In order to calculate φ in terms of λ explicitly, we assume the solution for q asfollows.

q = B cos(λt + Φ) = B cos(Ωt + Φ). (13)

Here B and Φ are to be determined. Consequently, p can be found as follows.

p = q = −BΩ sin(Ωt + Φ) = −Ω√

B2 − q2. (14)

Thus, the G in Eq. (12) can be calculated as follows.

G = pdp

dq= (−Ω

√B2 − q2)(−Ω

12

1√B2 − q2

(−2q)) = Ω2q. (15)

When the instability occurs, dq/dq = ∞. The instability occurs when the followingequation is satisfied.

∂2H

∂q2− ∂N

∂q+

∂G

∂q= 0, or ω2 − Ω2 = 0. (16)

Clearly, Eq. (16) suggests the system is unstable only when the driving frequency isequal to the natural frequency of the system, as expected.

The above results can be generalized to the problem with many degrees of freedom.With external excitations, a system becomes unstable when the driving frequency co-incides with one of the natural frequencies of the system. However, the system can bestabilized in the sense that no unbounded responses occur when damping elements areincluded. Without driving force, the system is internally stable when both mass andstiffness are positive or negative. Following the similar spirit in Eqs. (2)–(5), we explorethe stability of the 2-dof viscoelastic models, shown in Figure 2, with negative stiffness el-ements under external excitations, and cross-link our results with the Lyapunov indirecttheorem.

3. Equations of motion. The mechanical system considered here is shown as Figure2(a), in which the left module is called element 1, and the right one is called element 2.Figure 2(b) is a simplified version of Figure 2(a), intended to be used in later parametricstudy. The subscripts indicate to which module a component belongs. The mass pointsm1 and m2 are called node 1 and node 2, respectively. Following Newton’s second law,the equations of motion of the mechanical system can be expressed as follows.


m1η1

y1

u1, F1

z1

k1

c1

k2

c2 m2

u2, F2

y2

η2

z2

m1η

y

u1, F1

z

k1

k2

m2

u2, F2

c1

Fig. 2. (a) The 2-dof viscoelastic model. (b) The simplified versionof (a), used in parametric study.


[m1 0

0 m2

](u1

u2

)+

[c1 + c2 −c2

−c2 c2

](u1

u2

)+

[k1 + k2 −k2

−k2 k2

](u1

u2

)+(

f1 − f2

f2

)=(

F1

F2

)(17)

where

f1 +γ1c1

y1 + α1y1f1 =

α1y21

y1 + α1y1u1 +

y1γ1c1

y1 + α1y1u1, (18)

f2 +γ2c2

y2 + α2y2f2 =

α2y22

y2 + α2y2(u2 − u1) +

y2γ2c2

y2 + α2y2(u2 − u1), (19)

are the constitutive equations for the internal force and deformation in the standardlinear solids (i.e., the y1 − η1 − z1 or y2 − η2 − z2 module). The symbols f1 and f2

denote the internal force in the standard linear solid elements. Also, we define yi = κiki,ηi = γici and zi = α2yi, where i = 1, 2, to facilitate our later parametric study.

3.1. Solutions in the frequency domain. Investigating the solutions of the equations ishelpful in understanding the response of the systems at different frequency. After Fouriertransform, the governing equations of the system, Eqs. (17)–(19), can be converted intothe following algebraic equations.

T(

u1

u2

)=

(F1

F2

), (20)

where

T =[ −ω2m1 + k1 + k2 + iω(c1 + c2) + d1 + d2 −(k2 + iωc2) − d2

−(k2 + iωc2) − d2 −ω2m2 + k2 + iωc2 + d2

], (21)

d1 =α1y

21 + iωy1γ1c1

y1 + α1y1 + iωγ1c1, (22)

d2 =α2y

22 + iωy2γ2c2

y2 + α2y2 + iωγ2c2. (23)

It is noted that the determinant of the coefficient matrix, Eq. (21), dominates theboundedness of the displacement responses in frequency domain. In other words, if thecoefficient matrix becomes singular, the system will be unstable in the sense that finiteinput produces unbounded output. However, it is not required for T to be positivedefinite for stability. We remark that for gyroscopic systems, T is not symmetric, asdiscussed in [1], [2], [7] and references therein. One usually encounters gyroscopic systemswhen follower forces are considered. Our system is not gyroscopic. The effective complexcompliance (jeff ) and effective stiffness (keff ) can be calculated at a specified frequencyas follows.

jeff = j′eff + j′′eff =u2

F2

, and (24)

keff = k′eff + k′′

eff =F2

u2. (25)


The effective tan δ is defined as follows.

tan δ =k′′

eff

k′eff

, or tan δ = − j′′eff

j′eff

. (26)

For the special case m1 = m2 = 0, c1 = c2 = 0, y2 = 0, k1 = 10, and k2 = 5 kN/m,the overall tan δ can be obtained from tan δ = h/g, where

h(y, z, η, ω) = 5ωηy2A2, (27)

g(y, z, η, ω) = (ω2η2 + z2)(ω2η2 + (y + z)2) + 25y(ω2η2 + z(y + z))A2 + 150y2A4, (28)

A =∣∣∣∣1 +

z

y+ i

ωη

y

∣∣∣∣ . (29)

Here z = αy. Choosing zero masses is done to reduce some mathematical complexities,and at the same time, simulates a composite material in the continuum sense. This isthe special case studied in [13]. Later, we will discuss some interesting singular behaviorof the overall tan δ for this special case.

3.2. Solutions in the time domain. The general solution of Eq. (17) in the time do-main is quite complicated. For the purpose of demonstrating extreme properties of themechanical system due to a negative stiffness element, we assume c2 = 0, κ2 = 0, andγ2 = 0, as shown in Figure 2 (b). One should be aware that artificially setting materialproperties to be zero causes certain degrees of degeneracy in solutions. We will pointout the effects of degeneracy along with our derivation. The governing equation can beexpressed as follows.[

m1 00 m2

](u1

u2

)+

[c1 00 0

](u1

u2

)+

[k1 + k2 −k2

−k2 k2

](u1

u2

)+(

f1

0

)=(

F1

F2

). (30)

To decouple the above equation, one multiplies the inverse of the stiffness matrix (non-singular stiffness matrix is assumed) on the both sides of the equation, and then obtainsthe following.[

m1k1

m2k1

m2k1

k1+k2k1k2

m2

](u1

u2

)+

[c1k1

0c1k1

0

](u1

u2

)

+[

1 00 1

](u1

u2

)+

(f1k1f1k1

)=

(F1k1

+ F2k1

F1k1

+ k1+k2k1k2

F2

).(31)

It can be seen that the above equation is not fully decoupled yet. However, if onemakes a further assumption that m1 = m2 = 0, the equations can be decoupled asfollows.

u1 +k1

c1u1 +

f1

c1=

1c1

(F1 + F2), (32)

u2 +k1

c1u2 +

f1

c1=

F1

c1+

k1 + k2

c1k2F2, (33)


where

f1 = e−y1(1+α1)t

η1

[∫ t

0

(α1y

21

γ1c1u1 + y1u1)e

y1(1+α1)tη1 dt + f1(0)

]. (34)

The zero mass assumption is implemented in later numerical study by assigning m1 =m2 = 10−8. Also, the assumption is legitimate in our study since the analysis is intendedto model composites in the continuum sense. Note again, by doing so, m-degeneracyis unavoidable. Appendix A shows the effects of the m-degeneracy through a 1-dofexample. Eq. (34) is the solution of Eq. (18). f1(0) is the initial condition, equalto y2u1(0). Observing Eq. (33), it can be found that the solution of u2 is completelydetermined by that of u1, as in Eq. (35). From now on, we assume F1 = 0 throughoutthe rest of the analysis. The physical rationale is to simulate the mechanical behaviorof two phase composites, in which the interface between the two phases has no externalforce applied independent of the solid phases.

u2 = e−k1t

c1

[∫ t

0

(k1 + k2

c1k2F2 − f1

c1)e

k1tc1 dt + u2(0)

]. (35)

As for the solution of u1, one introduces Eq. (34) into Eq. (32), and differentiatesboth sides of the equation with respect to time to eliminate the integral from f1.

c1u1 + (k1 + y1 +y1(1 + α1)

γ1)u1 + (

k1y1(1 + α1)γ1c1

+α1y

21

γ1c1)u1 =

y1(1 + α1)γ1c1

F2 + F2. (36)

Eq. (36) is a second-order constant coefficient ordinary differential equation with non-homogeneous (or forcing) terms. Let F2 = P2 + A2 cosΩ2t; one can find the generalsolution as follows. P2 and A2 are pre-chosen constants.

u1 = uh1 + up

1, (37)

uh1 = C1e

λ1t + C2eλ2t, (38)

where

λ1,2 =−(k1 + y1 + y1(1+α1)

γ1) ±

√(k1 + y1 + y1(1+α1)

γ1)2 − 4(k1y1(1+α1)

γ1+ α1y2

1γ1

)

2c1, (39)

up1 =

y1(1 + α1)γ1c1

[1

λ1λ2P2 +

A2

Ω22 + λ2

2

(λ1λ2 − Ω2

2

Ω22 + λ2

1

cosΩ2t +(λ1λ2)Ω2

Ω22 + λ2

1

sin Ω2t)]

− 1(D − λ1)(D − λ2)

F2. (40)

Here, D is the differential operator, defined as d/dt. In the solution, C1 and C2

are constants to be determined by initial conditions, and the particular solution can becarried out more explicitly. However, it is enough for stability discussion.

Observe that if c1 = 0, the solution of Eq. (36) is

u1 = e−k1y1(1+α1)−α1y2

1η1(k1+y1)

[∫ t

0

F2

k1 + y1e

k1y1(1+α1)+α1y21

η1(k1+y1) dt + u1(0)

], where k1 + y1 = 0.

(41)


Consequently, following Eq. (33),

u2 =k1 + k2

k1k2F2 − f1

k1, (42)

where f1 is determined by Eq. (34). It noted that when k1 + y1 = 0, u1 becomesunbounded, and so does u2 consequently. Based on the solutions for the displacements,Eqs. (34) and (41), one can define two time constants, as follows.

τ1 =η(k1 + y)

k1y(1 + α) + αy2, for u1. (43)

τ2 =η

y(1 + α), for u2. (44)

It is noted that Eq. (43) is the relationship between the time constant and the rate ofoverall divergence for the system, as discussed in [14] by using the Lyapunov indirectmethod. As for c1 = 0, the solutions of Eqs. (32) and (33) are more complicated, andcan be shown as follows, when F1 = 0.

u1 = C1eλ1t + C2e

λ2t + up1, and (45)

u2 = e−k1t

c1

[∫ t

0

(k1 + k2

c1k2F2 − f1

c1)e

k1tc1 dt + u2(0)

], (46)

where

λ1,2 =−(k1 + y1 + c1y1(1+α1)

η ) ±√

(k1 + y1 + c1y1(1+α1)η )2 − 4c1(

k1y1(1+α1)η + α1y2

1η )

2c1,

(47)

f1 = e−y1(1+α1)

η

[∫ t

0

(α1y

21

γ1c1u1 + y1u1)e

y1(1+α1)η dt + f1(0)

]. (48)

Similarly, time constants for the solutions can be defined as follows.

τ0 = − 1λ1

and τ1 = − 1λ2

, for u1, (49)

τ2 =η1

y1(1 + α1), for u2. (50)

If the time constant is positive, the solution, corresponding to the relevant degree offreedom, is exponentially decaying. Negative time constants indicate instability. Theapplied force (F2) and particular solution (up

1) do not change the boundedness of thesolutions if F2 is a bounded function with respect to time.

4. Stability analysis. It is understood that the Lyapunov indirect method, orRouth-Hurwitz criterion, for stability analysis is based on a first order approximation. Inother words, if a system fails the stability test with the Lyapunov indirect method, theresponse of the system is unbounded under infinitesimal perturbation. However, the rateof divergence may be controllable. For the mechanical system, as shown in Figure 2(b),one needs to find the eigenvalues of the Jacobian matrix of Eq. (51), which is derivedfrom Eqs. (17)–(19) through the state-space technique [8] with κ2 = 0 and γ2 = 0. Then,


by tuning the parameters y, η, and α, we identify that the regimes containing eigenvalueswith positive real part are unstable. α is a dimensionless parameter, defined as α = z/y.

u1

u2

v1

v2

f

=

0 0 1 0 00 0 0 1 0

−k1+k2m1

k2m1

− c1+c2m1

c2m1

− 1m1

k2m2

− k2m2

c2m2

− c2m2

0αy2

η 0 y 0 − y(1+α)η

u1

u2

v1

v2

f

+

00F1m1F2m2

0

. (51)

However, since the solutions for the system, shown in Figure 2 (b), have been explicitlyderived, as expressed in Eqs. (41) and (42) for c1 = 0, or Eqs. (45) and (46) for c1 = 0,one can examine the stability of the system via the boundedness of the solutions as timeapproaches infinity. Direct investigating the boundedness of the solutions is equivalentto Lyapunov indirect method for stability analysis. The major benefit of investigatingthe solutions in the time domain is that the stability analysis can be done in a moretransparent manner. Later, in our parametric study, we directly solve the eigenvaluesof the Jacobian matrix in Eq. (51). It has been identified that the time constant inEq. (43) is responsible for the only eigenvalue with positive real part in the Jacobianmatrix, as discussed in [14]. In the following, we derive stability criteria on the physicalquantities of our model from the solutions in the time domain. First, for c1 = 0, fromEq. (46), the conditions for u2 to be bounded are as follows.(a) k1/c1 > 0, so that there will be no exponentially growing behavior due to u2(0).(b) f1(t) and F2(t) need to be non-exponentially growing functions, such as constant,trigonometric functions, or exponentially decaying functions.

In contrast, when c1 = 0, from Eq. (42), Condition (a) does not exist for u2(t) to bebounded. Furthermore, even in the case c1 = 0, since we assume that k1 > 0 and c1 > 0throughout, Condition (a) is trivial. Since F2 is the only applied force, one can artificiallyset it to satisfy Condition (b). The behavior of f1(t) needs further investigation. FromEq. (34), it can be determined that the following conditions must satisfy Condition (b).(c) y > 0 and α > −1, or y < 0 and α < −1, so that there will be no exponentiallygrowing behavior due to f1(0).(d) u1(t) and u1(t) must be non-exponentially growing functions.

In order to satisfy Condition (d), one needs to look into the solution of u1, Eq. (45),carefully. It can be understood that the particular solution of u1 will be bounded if F2

is bounded and the driving frequencies are not equal to the resonant frequencies of thesystem (rigorously speaking, it is not the conventional resonant frequencies (≈√K/M),but λ1 and λ2; see Eq. (47)). As for the homogeneous solution of u1, one can derive thestability conditions from Eq. (47) as follows, with the definition of y = κk. One of thefollowing two conditions is enough to ensure the stability.(e) 1 + κ + κ(1+α)

γ > 0 and (1 + κ + κ(1+α)γ )2 − 4(κ(1+α)

γ + ακ2

γ ) < 0.

(f) 1 + κ + κ(1+α)γ > 0 and (κ(1+α)

γ + ακ2

γ ) > 0.For the second equation in Condition (e), we have checked that α does not have

purely real-number solutions in κ = −100 to 100, for γ = 1, 10, 100, 1000. Therefore,we conjecture that the inequality cannot be satisfied for a wide range of α, κ, and γ.


Consequently, Condition (e) will not be satisfied. Further analysis shows that Condition(f) can be replaced by the following two conditions.(f1) κ > 0 and α > −γ

κ − γ − 1, or κ < 0 and α < −γκ − γ − 1, for the first inequality in

(f).(f2) κ > 0 and ακ +α + 1 > 0, or −1 < κ < 0 and α < − 1

1+κ , or κ < −1 and α > − 11+κ ,

for the second inequality in (f).Conditions (f1) and (f2) have to be satisfied simultaneously to fulfill Condition (f). In

contrast, when c1 = 0, from Eq. (35), Condition (e) is not necessary, and Condition (f)must be replaced by the following one.(g) ky(1+α)+αy2

η(k+y) > 0.Condition (g) can be further simplified as follows, with the definition of y = κk, when

k/η > 0.(g’) 1 + κ > 0 and ακ2 + ακ + κ > 0, or 1 + κ < 0 and ακ2 + ακ + κ < 0.

It can be shown that Condition (g’) is equivalent to Condition (f2). In the case fory < 0 (negative stiffness in the standard linear solid element), Conditions (c)2 and (e), orConditions (c)2, (f1), and (f2) must be satisfied to ensure bounded responses, as shownin Figure 3, the shaded area as an example of stability regimes with a negative stiffnesselement. The first quadrant of Figure 3 also indicates stability, corresponding to positivestiffness for all the elements.

5. Discussion. It is noted again that the parametric study here and in the followingis based on the model shown in Figure 2(b). The stability map, as shown in Figure 3,is valid as long as the driving force, F2, is a non-exponentially growing function and itsdriving frequency is not the natural frequency of the system. The dotted line indicatesthe asymptote of the stability boundary. From the stability analysis above, one canidentify the stability regions as follows.(I) κ > 0 and α > 0 (not interesting, since positive stiffness for all the elements),(II) −1 < κ < 0 and α < − 1

1+κ , and(III) κ < −1 and α > − 1

1+κ , depending on γ.The shaded area is the stable region for any γ. For the c-degenerate case, i.e., c = 0, γ

must be set as large as possible to maintain finite η, the damping in the standard linearsolid. In this case, case (III) does not depend on γ any more. However, it does notmean that the region, α > − 1

1+κ , is stable because of Condition (c). This result impliesthat if the response of the system at node 1 is stable, i.e., bounded, then that at node2 is stable, as well. Thus, the trajectories of the symbols, solid square and open circle,coincide. One is the stability boundary and the other is the trajectory causing infinitecompliance with c = 0 and η = 0 for the 1-dof case, i.e., k2 = 0, η2 = 0 (see Eq. (63) inAppendix B). As will be seen later, to achieve extreme high stiffness, one needs to be inthe region where κ < −1.

Figure 4(a) shows the quasi-static response of the model in Figure 2(b) with respect tothe negative stiffness element (y). The highest compliance at y = −1.6 kN/m is reachedsimultaneously for nodes 1 and 2 due to stiffness neutralization in Element 1. The lowestcompliance (i.e., highest stiffness) occurs at about y = −2.6 kN/m. The real part of thecompliance of the system at node 1 is negative when y < −1.6 kN/m, and that at node


Fig. 3. Stability map for α, γ, and κ, based on the solutions ofgoverning differential equations in time domain. The shaded regionand the first quadrant (i.e., κ > 0 and α > 0) correspond to stability.

2 is negative when −2.6 < y < −1.6 kN/m. The stability analysis based on Lyapunov’stheorem shows a stable regime, as indicated in Figure 4(b). Since α is negative, the role ofthe negative stiffness element switches between the y- and z-springs, again by definitionz = αy. Thus, the system becomes unstable when y > 0 and y < −1.6 kN/m. Theparameters η and ω are assumed to be 0.0001 kN-s/m and 0 rad/s, respectively, in bothFigures 4 and 5. It is noted that for such a small viscosity η, a standard linear solid withno negative stiffness elements has a time constant (τ ≈ viscosity/stiffness) on the orderof 10−5 seconds, if the stiffness of spring elements is about 10 kN/m. Consequently, theDebye peak for the tan δ of the standard linear solid will be at a frequency of about 105

rad/s. In Figure 5, we plot the compliance curves and the stability-losing eigenvalue withrespect to the negative stiffness (y), for different α′s. It can be seen that the spacingbetween extreme high compliance and stiffness can be reduced by increasing α. But,the distance cannot be reduced arbitrarily much, as can be seen for large α′s. Also, thestability-losing eigenvalue has larger magnitude, indicating higher rate of divergence, i.e.,more unstable, for large α. It is noted again that the stability boundary from previousanalysis is at extreme high compliance.


(a)

(b)

Fig. 4. Quasistatic behavior: no inertial terms. (a) Compliance,

(u1/F2) and (u2/F2) , on a linear scale (left) and absolute com-

pliance, |u1/F2| and |u2/F2| , on a logarithmic scale (right) vs. ywith ω = 0 and α = −1.2. (b) Overall compliance and the stability-losing eigenvalue vs. y with ω = 0 and α = −1.2. We define z = αy,where y and z are stiffness in Figure 2 (b) with c1 = 0. The unstablebranch for y > 0 is due to α < 0.


Fig. 5. Quasistatic behavior: no inertial terms. Effective compliance(|jeff |) vs. negative stiffness (y) in the purely elastic limit with

different α′s. k1 = 10, k2 = 5 kN/m, m1 = m2 = 10−8 kg, η =0.0001 kN-s/m and ω = 0 rad/s. We define z = αy, where y andz are stiffness in Figure 2 (b). For clarity, compliance curves areseparated by multiplying by a constant.

In Figures 6, 7, and 8, we show the high frequency responses of the system under theinfluence of negative stiffness with η = 0.0001 kN-s/m and α = −5. All peaks are struc-tural resonant and anti-resonant frequencies. The purpose of studying high frequencyresponses is to understand the process of neutralization due to negative stiffness. Figure 6shows the frequency response of compliance at discrete negative stiffness in y. It is notedwhen y = −15 kN/m, there is no anti-resonant-like behavior. In Figure 7, the peaks arealso structural resonances for high frequency responses. It shows that the anti-resonantpeaks (high stiffness) move to the low negative stiffness regime (i.e., in the stable zone,if negative stiffness is low enough), as driving frequency increases. Excessive negativestiffness makes anti-resonances disappear. From Figures 6 and 7, it also can be seenthat by tuning either frequency or negative stiffness, one can alter the spacing betweenresonant and anti-resonant peaks, or even flip their order of appearance with respect tofrequency or negative stiffness. We emphasize that at the low frequency limit the up-ward or downward peaks in compliance are not structural resonant phenomena, but athigh frequency the peaks indicate the natural frequencies of the system. The shift in thenatural frequency is due to cancellation in overall stiffness. The stability analysis of the


10-10

10-8

10-6

10-4

10-2

100

102

104

106

108

103 104 105

Abs

(Com

plia

nce)

, | j

|, (k

N/m

)-1

Frequency, rad/sec

| j | x 106, y=0

| j | x 104, y=-1.2

| j | x 1, y=-5

| j | x 10-3, y = -10

| j | x 10-6, y = -15

Node2

Node1

Node2

Node2

Node2

Node2

Node1

Node1

Node1

Node1

Fig. 6. Dynamic compliance vs. frequency with different negativestiffness (y) in units of kN/m. m1 = m2 = 10−8 kg. k1 = 10, k2 = 5kN/m, α = −5, η = 0.0001 kN-s/m. Absolute compliances at node

1 and node 2 are calculated from |u1/F2| and |u2/F2|, respectively.The symbol |j| denotes compliance either at node 1 or node 2. Basedon Figure 2(b) with c1 = 0. The upward and downward peaks arestructural resonances and anti-resonances, respectively. Thin solidlines are the compliance at node 1 and thick ones are the effectivecompliance of the system. For clarity, curves are separated by mul-tiplying by a constant.

system driven at 25k rad/sec is shown in Figure 8, with respect to negative stiffness. Itshows that the extreme high stiffness (y = −7 kN/m) is located in the stable regime. Theresult is not too surprising physically, because one can easily obtain extreme dynamicalstiffness due to the structural anti-resonant response and negative stiffness plays a role toshift the structural anti-resonance, but the methodology is based on Lyapunov’s indirecttheorem. Again, the instability in the regime y > 0 is due to α < 0. When y < −7 ory > 0, eigenvalues split. The eigenvalue whose real part is greater than zero causes thesystem to be unstable, as discussed in [12].

In Figure 9 (a), we plot the h and g function in Eqs. (27) and (28) to demonstratethe singularities in tan δ. This phenomena have been reported in [13] and [14]. Here, we


Fig. 7. Dynamic compliance vs. y with different frequencies. Fre-quency is in units of 103 rad/sec. m1 = m2 = 10−8 kg. k1 =10, k2 = 5 kN/m, α = −5, η = 0.0001 kN-s/m. Absolute compli-

ances at node 1 and node 2 are calculated from |u1/F2| and |u2/F2|,respectively. The symbol |j| denotes compliance either at node 1 ornode 2. Based on Figure 2(b) with c1 = 0. Peaks indicate struc-tural resonances for high frequency responses. For clarity, curves areseparated by multiplying by a constant.

mathematically investigate the behavior of tan δ under the influence of negative stiffness.It is noted that in this analysis, we set z = 5 kN/m and ω = 1 rad/sec. As seen, when η

is small (less than about 0.4 kN-s/m), there are two zeros in the g function, indicatingtwo singularities in tan δ, since the h function is finite, in the negative stiffness rangeplotted. When η is large, the singularities in tan δ disappear and the tan δ curve appearsas a hump, as reported in [13] and [14]. In Figure 9 (b), we show that the stabilityboundary, calculated based on Lyapunov’s indirect theorem, coincides at the right handside of the first zero (from the origin) in g. It indicates that the singularities of tan δ arelocated in the unstable (or metastable) regime. However, in a limiting case, as η → 0,one can achieve stable singular tan δ.

6. Conclusions. We rigorously derive the stability criteria for the 2-dof viscoelasticsystem containing a negative stiffness element. The stability conditions derived from thetime domain solutions of the differential systems are good for the response of the system


Fig. 8. Routh-Hurwitz eigenvalues and dynamic compliance vs. y,at ω = 25k rad/sec. m1 = m2 = 10−8 kg. k1 = 10, k2 = 5 kN/m,α = −5, η = 0.0001 kN-s/m. Based on Figure 2(b) with c1 = 0. Theunstable branch for y > 0 is due to α < 0.

at any frequency. Stable extreme stiffness at high frequency is demonstrated. Stableextreme tan δ at low frequency is verified. In the low frequency limit, the system showsstable extreme high compliance and metastable extreme high stiffness.

7. Appendix A: A case of m-degeneracy. The purpose of this appendix is todemonstrate the effects of c-degeneracy and m-degeneracy. We consider the 1-dof vis-coelastic model, as shown in Figure 10. The governing equation of the system can beexpressed in the state-space representation, as follows. u

v

f

=

0 1 0− k

mcm − 1

mαy2

η y − y(1+α)η

u

v

f

+

0Fm

0

. (52)

It is understood that the characteristic equation of the Jacobian matrix in Eq. (52) is acomplicated third-order polynomial. The solutions of the characteristic equation are theeigenvalues for the Lyapunov stability analysis. However, for c = 0 (c-degeneracy), one


10-1

100

101

102

103

104

105

-20

-15

-10

-5

0

5

10

15

20

-5 -4.5 -4 -3.5 -3

h(y,

z, η

, ω)

g(y, z, η, ω

)

y, kN/m

η=0.01

η=0.3

η=0.5

η=0.9

η=0.9

η=0.5

η=0.3 η=0.1

η=0.1

η=0.01

-20

-15

-10

-5

0

5

10

15

20

-20

-15

-10

-5

0

5

10

15

20

-3.8 -3.7 -3.6 -3.5 -3.4 -3.3 -3.2

g(κ 1, κ

2, η, ω

)R

e(Eigenvalues), sec

-1

y, kN/m

StableUnstable

Fig. 9. Singularities in tan δ and their stability, calculated with k1 =10, k2 = 5, z = 5 kN/m, ω = 1 rad/s. Based on Figure 2 (b) withc1 = 0. (a) Singularities in tan δ. g = 0 indicates singularities. (b)Stability of the singularities in tan δ. Curved lines are calculated gfunction with respect to various η, defined in (a). Stability analysisis based on Eq. (51).


mη

y

u, F

z

k

c

Fig. 10. One-dof viscoelastic model.

can write down the simpler characteristic equation as follows.

λ3 +y(1 + α)

ηλ2 + (

k

m+

y

m)λ +

αy2

mη+

y(1 + α)mη

= 0. (53)

However, if one further assumes m = 0 (m-degeneracy), the characteristic equationbecomes

λ =k(ακ2 + ακ + κ)

η(1 + κ). (54)

As seen, from Eq. (53) to (54), the numbers of roots of the characteristic decrease from3 to 1, which means that there is some information missing for stability analysis due todegeneracies.

8. Appendix B: Further demonstration through a 1-DOF viscoelasticmodel. The purpose of this appendix is to show the statement made in the first para-graph in the discussion. Consider the mechanical model shown in Figure 10. The corre-sponding governing equations in time domain are as follows, when m = 0.

ku + cu + f = F, (55)

f +η

y + zf =

yz

y + zu +

yη

y + zu, (56)

where y = κk, z = ακk, and η = γc. The symbol f denotes the internal force in thestandard linear element. Through Fourier transform, the governing equations in the


frequency domain are

(k + icω)u + f = F , (57)

(y + z + iωη)f = (yz + iωyη)u. (58)

One can calculate the effective compliance as follows.u

F=

A

A2 + B2− i

B

A2 + B2, (59)

where

A = k + k1α(1 + α) + ω2β2

(1 + α)2 + ω2β2, (60)

B = ωc(1 +γ

(1 + α)2 + ω2β2), and (61)

α = k2/k1, β = η/y, γ = η/c. (62)

For c = η = 0, i.e., B = 0, the extreme (infinite) compliance occurs at A = 0, i.e.,

y = −1 + α

α. (63)

For a fixed k, one can draw two hyperbolic curves on the α − y plane. To derive thesolution of Eq. (55) and (56) in time domain directly, we first rewrite the governingequation in the following way.

cu + (k + y +y(1 + α)

γ)u + (

ky(1 + α)γc

+αy2

γc)u =

y(1 + α)γc

F + F . (64)

The general solution of Eq. (64) is as follows.

u = uh + up. (65)

The homogeneous solution is

uh = C1eλ1t + C2e

λ2t, (66)

where C1 and C2 are determined by initial conditions and

λ1,2 =−(k + y + y(1+α)

γ ) ±√

(k + y + y(1+α)γ )2 − 4(ky(1+α)

γ + αy2

γ )

2c. (67)

The positive real part of λ′s indicates that the homogeneous solution, uh, becomes un-bounded as time increases, which is unstable in the sense of Routh-Hurwitz. In order toobtain stable (non-growing) solutions, one of the following two sets of inequalities mustbe satisfied, for k > 0. Let y = κk and c = 0.(B-1) 1 + κ + κ(1+α)

γ > 0 and (1 + κ + κ(1+α)γ )2 − 4(κ(1+α)

γ + ακ2

γ ) < 0 . It is conjecturedthat the second inequality will not hold for a wide range of α, γ, and κ.(B-2) 1 + κ + κ(1+α)

γ > 0 and (κ(1+α)γ + ακ2

γ ) > 0.As seen, Condition (B-2) here is the same as Condition (f) in the 2-dof case. Figure

11 shows the compliance curves, based on Eq. (59), and the results from Lyapunov’sstability analysis. It shows that if the amount of negative stiffness is more than thatcorresponding to the highest compliance, the system will be unstable. Since negative α

is used in the graph, there is another unstable branch when y is sufficiently positive.


Fig. 11. Compliance and eigenvalue analysis of the 1-dof system inFigure 10.

References

[1] R. Bulatovic, On the Lyapunov stability of linear conservative gyroscopic systems, C. R. Acad. Sci.Paris, t. 324, Serie II b, p. 679-683 (1997).

[2] P. Gallina, About the stability of non-conservative undamped systems, Journal of Sound and Vi-bration, 262, 977-988 (2003). MR1971816 (2004b:70044)

[3] R. S. Lakes, “Extreme damping in composite materials with a negative stiffness phase”, Physical

Review Letters, 86, 2897-2900 (26 March 2001).[4] R. S. Lakes, “Extreme damping in compliant composites with a negative stiffness phase” Philosoph-

ical Magazine Letters, 81, 95-100 (2001).[5] R. S. Lakes, T. Lee, A. Bersie, and Y. C. Wang, “Extreme damping in composite materials with

negative stiffness inclusions”, Nature, 410, 565-567 (29 March 2001).[6] R. S. Lakes and W. J. Drugan, Dramatically stiffer elastic composite materials due to a negative

stiffness phase?, J. Mechanics and Physics of Solids, 50, 979-1009 (2002).[7] H. Leipholz, Stability of elastic systems, Sijthoff & Noordhoff, Alphen aan den Rijn, The Nether-

lands, 1980. MR0595163 (82g:73001b)[8] L. Meirovitch, Methods of Analytical Dynamics, McGraw-Hill, New York, 1970.[9] J. W. Morris Jr. and C. R. Krenn, The internal stability of an elastic solid, Philosophical Magazine

A, 80, 12, 2827-2840 (2000).

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[10] S. H. Strogatz, Nonlinear Dynamics and Chaos, Perseus Books, Cambridge, Massachusetts, 1994.[11] Y. C. Wang and R. S. Lakes, Extreme thermal expansion, piezoelectricity, and other coupled field

properties in composites with a negative stiffness phase, J. Appl. Phys. 90, 6458-6465 (2001).[12] Y. C. Wang and R. S. Lakes, Extreme stiffness systems due to negative stiffness elements, American

Journal of Physics, 72, 40–50 (2004a).[13] Y. C. Wang and R. S. Lakes, Stable extremely high-loss discrete viscoelastic systems due to negative

stiffness elements, Appl. Phys. Lett. 84, 4451–4453 (2004b).[14] Y. C. Wang and R. S. Lakes, Negative Stiffness Induced Extreme Viscoelastic Mechanical Properties:

Stability and Dynamics, Philos. Mag., accepted (2004c).

Documents

STABILITY OF NEGATIVE STIFFNESS VISCOELASTIC SYSTEMSsilver.neep.wisc.edu/~lakes/NegStfQAM05.pdf · with extreme high stiﬀness analyzed here was metastable. We established an explicit